C1

Ch1 Algebra and

functions

Quadratic functions

Equations and Inequalities

Sketching curves

Coordinate geometry

Sequences and Series

Differentiation

Integration

2

Surds Some numbers like √2, √3 cannot be written as a decimal because they go on forever. Numbers like this are called surds. For example √2 = 1.4142 . . .

3

You can manipulate surds by using the rule √(a X b) = √a X √b and vice versa √a X √b = √(a X b)

√12 = √(4 X 3)

= √4 X √3 = 2 X √3 = 2√3

√(4 X 5) = √4 X √5 √3 X √7 = √(3X7)

4

5√6 – 2√24 + √294 = 5√6 – 2√(4X6) + √(49X6)

= 5√6 – 2 X √4 X √6 + √49 X √6 = 5√6 – 2 X 2 X √6 + 7 X √6 = 5√6 – 4√6 + 7√6 = 8√6

5

Another rule we can use to manipulate surds is

�ab =

√a√b

or vice versa

√a√b

= �𝑎𝑎𝑏𝑏

√122

= √(4X3)

2

= √4 X √3

2

= 2 X √3

2

= √3

6

√44√11

= �4411

= √4 = 2

7

If we have a fraction with a surd on the bottom like 4

√11 we can rationalise the denominator. This means we

turn the bottom or denominator into a rational or non surd. We might do this if we get a surd on the bottom of our answer at the end of our question. This just tidies the fraction up and it is better to give an answer in this way.

1√3

multiply top and bottom by √3

= 1 X √3

√3 X √3

= √33

This is exactly the same number but now the surd is on the top instead of the bottom.

8

1

2√5

multiply top and bottom by √5

= 1 X √5

2√5 X √5

= √5

2 X 10

= √520

This is exactly the same number but now the surd is on the top instead of the bottom so it is tidier.

11 + √2

multiply top and bottom by (1 - √2)

= 1 X (1 - √2)

(1 + √2)(1 - √2)

= 1 - √2

1 + √2 - √2 + 2

= 1 - √2

3

Notice that this works because multiplying (1 + √2) by (1 - √2) we end up with +√2 and -√2 which cancel out.

9

11 + √2

multiply top and bottom by (1 - √2)

= 1 X (1 - √2)

(1 + √2)(1 - √2)

= 1 - √2

1 + √2 - √2 + 2

= 1 - √2

3

Notice that this works because multiplying (1 + √2) by (1 - √2) we end up with +√2 and -√2 which cancel out.

10

Chapter 2 – Quadratic equations Quadratic equations are equations with ‘x2’s in.

An equation like 2x + 3 = 11 (a linear equation) has only one answer that makes it true (x = 4).

Quadratic equations often (but not always) have two solutions that make them true.

For example, take x2 – 5x + 6 = 0, there are two roots, substituting x = 2 or x = 3 works.

x2 – 5x + 6 22 – 5 X 2 + 6 = 0 √ 32 – 5 X 3 + 6 = 0 √

Notice that putting in any other number is not going to get us zero and it is therefore not a root or solution.

There are three ways to solve quadratic equations

- Factorisation

- Completing the square

- Using the formula.

11

Factorisation Factorising means ‘solving by putting into brackets’. If we multiply out two brackets (x + 4)(x + 5) = x2 + 9x + 20 We notice that the number of ‘x’s is the same as the two numbers added together and the number by itself is the same as the two numbers added together. 4 + 5 = 9 4 X 5 = 20 This helps us when we go the other way and put a quadratic equation into brackets.

12

The other thing we need to realise for solving quadratic equations by factorisation is that if two things multiply together and equal zero then either the first thing or the second thing must equal zero. If

something X something else = 0 then either something = 0 or something else = 0

13

Solve the quadratic equation,

x2 – 5x + 6 = 0 --------------------------------------------------------------

Factorise (put into brackets) (x – 2)(x – 3) = 0 because (-2) + (-3) = -5 and (-2) X (-3) = +6 -------------------------------------------------------------- Either (x – 2) = 0 or (x – 3) = 0 so x = 2 or x = 3

14

Completing the square The second way of solving quadratic equations is by completing the square.

If we multiply out a bracket

(x + 3)2

= (x + 3)(x + 3)

= x2 + 6x + 9 -------------------------------------------------------------- (x + 3)2 = x2 + 6x + 9

Now if we take away 9 from both sides

(x + 3)2 – 9 = x2 + 6x or

x2 + 6x = (x + 3)2 - 9

Notice that the number in the bracket is half of the number in front of the x and the number by itself is half that number squared.

15

x2 + 8x = (x + 4)2 - 42

x2 + 8x = (x + 4)2 - 16

x2 - 10x = (x - 5)2 – (-5)2

x2 - 10x = (x - 5)2 - 25 Notice that it is – 25 not + 25!

x2 + 5x = (x + 52)2 – �5

2�

2

x2 + 5x = (x + 52)2 - 25

4

52 X 5

2 = 25

4

We can complete the square on quadratic expressions by looking at half the number in front of the x.

16

Solve the quadratic equation x2 + 8x + 10 = 0 by completing the square. -------------------------------------------------------------

x2 + 8x + 10 = 0 Check the coefficient of x2 equals one

x2 + 8x + 10 = 0 Move the constant to the other side

x2 + 8x = - 10 Complete the square on the expression on the left hand side.

(x + 4)2 – 42 = - 10

(x + 4)2 – 16 = - 10 Move the sixteen over to the other side

(x + 4)2 = - 10 + 16

(x + 4)2 = 6 Square root both sides and take away four

(x + 4) = +√6 or (x + 4) = -√6

x = - 4 + √6 or x = - 4 - √6

17

Solve the quadratic equation x2 - 12x + 7 = 0 by completing the square. --------------------------------------------------------------

x2 – 12x + 7 = 0 Check the coefficient of x2 equals one

x2 – 12x + 7 = 0 Move the seven to the other side

x2 - 12x = - 7 Complete the square on the expression on the left hand side

(x - 6)2 – 62 = - 7

(x - 6)2 – 36 = - 7 Move the 36 over to the other side and simplify

(x - 6)2 = - 7 + 36

(x - 6)2 = 29 Square root both sides and add six

(x - 6) = +√29 or (x - 6) = - √29

x = + 6 + √29 or x = + 6 - √29

18

Solve the quadratic equation 2x2 + 7x - 9 = 0 by completing the square. ------------------------------------------------------------------------

2x2 + 7x - 9 = 0 Make the coefficient of x2 = 1 by dividing all the way through by two

x2 + 72x - 9

2 = 0

Move the constant to the other side

x2 + 72x = + 9

2

Complete the square on the expression on the left hand side. Notice

that half of 72 is 7

4 .

(x - 74)2 – �7

4�

2 = + 9

2

(x - 74)2 – 49

16 = + 9

2

Move the 4916

over to the other side and simplify

(x - 74)2 = + 9

2 + 49

16

(x - 74)2 = 121

16

Square root both sides and move the 74 over

(x - 74) = + �121

16 or (x - 7

4) = − �121

16

x = 74 + �121

16 or x = 7

4 − �121

16

19

To complete the square 1. Make sure that the coefficient of x2 = 1, divide through if necessary. 2. Subtract the constant (number by itself) from both sides. 3. Complete the square for the quadratic expression. 4. Get all the numbers on one side and simplify. 5. Square root both sides.

20

Using the formula The third way of solving quadratic equations is by using the quadratic formula.

x =−b ± √ b2 − 4ac

2ac

Where a is the number in front of the x2, b is the number in front of the x and c is the number by itself. Notice that we will usually get two answers because of the ± bit.

Solve the quadratic equation, 3x2 – 7x + 11 = 0 --------------------------------------------------------------

a = + 3 b = -7 c = +11

21

Solve the quadratic equation x2 - 12x + 7 = 0 by using the quadratic formula.

x2 – 12x + 7 = 0

a = + 1 b = - 12 c = + 7 ------------------------------------------------------------------

x = −b ± √ b2−4ac 2ac

so

x = −b + √ b2−4ac 2ac

or x = −b − √ b2−4ac 2ac

x = −(−12) + �(−12)2 − 4.1.72.1.7

x = −(−12) − �(−12)2 − 4.1.72.1.7

x = (+12) + �(+144) − 2814

x = (+12) − �(+144) − 2814

x = (+12) + �(+116)14

x = (+12) − �(+116)14

x = 1.63 or x = − 0.09

22

A quadratic function will always have a distinctive parabola shape. We can tell which way up it will be when we draw it by looking at a, the number in front of the x2. f(x) = ax2 + bx + c

If a is positive then we will have a happy graph. Positive people SMILE!

If a is negative then we will have a sad graph. Negative people FROWN!

23

x =−b ± � (b2 − 4ac)

2a

We can work out how many times the graph will cut through the x axis and therefore how many roots we have by looking at the discriminant or the b2 – 4ac bit of the quadratic equation.

If b2 – 4ac is positive then we are adding and taking away a positive number so we get two different answers. The graph will cut the x axis at two different places and there will be two roots or solutions.

24

If b2 – 4ac equals zero then we are adding and taking away nothing so we get the same answer both times. The graph will just touch the x axis once and there will be only one root to this equation.

If b2 – 4ac is less than zero then we are trying to calculate the square root of a negative number which we can’t do, so the graph won’t go through the x axis at all and there will be no solutions to this equation.

25

Express the quadratic function f(x) = x2 + 4x + 9 in the form (x + A)2 + B. -------------------------------------------------------------- f(x) = x2 + 4x + 9

f(x) = {x2 + 4x} + 9 complete the square f(x) = {(x + 2)2 – 4} + 9 bring the numbers together f(x) = (x + 2)2 +5

Sketching quadratic functions Sometimes it can be useful to complete the square on quadratic functions especially if we are going to sketch them.

26

Write the quadratic function y = 2x2 - 3x - 2 in the form A(x + B)2 + C. -------------------------------------------------------------- f(x) = 2x2 – 3x - 2 take out a factor of 2

f(x) = 2[ x2 – 32x – 1 ]

f(x) = 2[ {x2 – 32x} - 1 ]

complete the square

f(x) = 2[ {(x - 34)2 – 9

16} – 1 ]

bring the numbers together

f(x) = 2[ (x - 34)2 – 9

16 – 1 ]

f(x) = 2[ (x - 34)2 – 25

16 ]

get into the form asked for

f(x) = 2(x - 34)2 – 50

16

27

This can help us when sketching quadratic graphs. Take y = x2 – 3x - 10, rearranging by factorising y = (x + 2)(x – 5) tells us that the curve cuts through the x axis at the points x = - 2 and x = + 5 (the roots). And rearranging by completing the square

y = (x - 32)2 - 49

4

tells us that the lowest the graph goes is − 494 = - 12.25 and

this happens when x = + 32 . The smallest anything squared

can be is zero, so the smallest y can be happens when the brackets equals zero. When the thing in the brackets is zero

y = − 494 and this happens when x = + 3

2 .

28

Chapter 3 – Equations and Inequalities Simultaneous equations are equations with two letters in like

2a + 3b = 7

3a – 2b = 4 There are an infinite number of solutions to just the first equation by itself, (a = 5 and b = -1) works as does (a = 0 and b = 7/3) but if we decide that our solution needs to work for both equations at the same time only one solution will work, (a = 2 and b = 1).

There are two ways to solve simultaneous equations Elimination and Substitution

29

Elimination To solve a simultaneous equation by elimination we need to multiply and get rid of one of the letters by adding or taking away the two equations.

Solve the simultaneous equation

2a + 3b = 8 2a + 3b = 8

3a – 1b = 23 X 3 9a – 3b =

69 11a = 77 1a = 7 ---------------------------------------------------------- Find the other letter by substituting back in to one of the original equations, it doesn’t matter which one. 2a + 3b = 8 2 X 7 + 3b = 8 14 + 3b = 8 3b = - 6 b = - 2

30

To solve a simultaneous equation by substitution we need to rearrange one of the equations till we get one letter by itself then substitute this into the other equation. Solve the simultaneous equation

2a + 3b = 8 3a – 1b = 23

------------------------------------------------------------------------ Rearrange the 2nd equation to get b by itself 1b = 3a – 23 ------------------------------------------------------------------------ Substitute 1b = 3a – 23 into the 1st equation 2a + 3b = 8 2a + 3(3a – 23) = 8 2a + 9a – 69 = 8 11a = 77 a = 11 ------------------------------------------------------------------------ Substitute (a = 11) back into one of the original equations to find the other letter. b = - 2

31

We can also use substitution to solve simultaneous equations where one is linear and the other a quadratic.

Solve the simultaneous equation

c + 2d = 3 c2 + 3cd = 10

--------------------------------------------------------------------------------------- Rearrange the 1st equation to get c by itself c = 3 – 2d --------------------------------------------------------------------------------------- Substitute c = 3 – 2d into the 2nd equation

c2 + 3cd = 10

(3 – 2d)2 + 3(3 – 2d)d = 10 --------------------------------------------------------------------------------------- Multiply out the brackets, add common things and bring together on one side so the d2 are positive. 2d2 + 3d + 1 = 0 This is a quadratic equation which can be factorised (put into brackets) and solved. (2d + 1)(d + 1) = 0

d = 12

d = 12 or d = - 1

32

Inequalities Find the set of values for which x2 – 4x + 5 < 0. Factorise it (put into two brackets) as normal. (x - 5)(x + 1) = 0 x = 5 and x = -1 are the critical values. Draw a sketch. it will be a happy graph because it is positive x2.

y = x2 - 4x - 5

The values for which x2 – 4x – 5 is less than zero is when it is below the x axis, i.e. when -1 < x < 4. Notice that this is just < not ≤ because the original question was only <.

33

Find the set of values for which x2 – 4x + 5 > 0. Find the critical values exactly the same as before. X2 – 4x + 5 = 0 (x + 1)(x – 4) = 0 The critical values are x = - 1 and x = +4.

y = x2 - 4x - 5

The graph is positive i.e. above the x axis when x < -1 or when x > 5.

34

Find the set of values for which 3 – 5x – 2x2 > 0. Find the critical values exactly the same as before. 3 – 5x – 2x2 = 0 (2x - 1)(x + 3) = 0

The critical values are x = 12 and x = -3.

If we sketch the graph out it will be a sad face because the x2 is negative.

y = 3 - 5x - 2x2

The graph is positive i.e. above the x axis when -3 < x

< 12.

35

Find the set of values for which 12 + 4x > x2 Rearrange and continue exactly as before. Find the critical values. 3 – 5x – 2x2 = 0 (2x - 1)(x + 3) = 0

The critical values are x = 12 and x = -3.

If we sketch the graph out it will be a sad face because the x2 is negative.

36

Chapter 4 – Sketching curves When we sketch a curve we do not need to draw it perfectly but just give an idea of the main shape and points of interest.

Some things we might be interested in are

- What is the general shape?

- Where does the graph cross the x axis?

- Where does it cross the y axis?

- Where are the maximum and minimum points?

- What happens as x gets very big?

- Are there any asymptotes?

37

Remember the general shape of a positive x3 + something curve.

And a negative x3 + something curve.

Notice they often cut through the x axis in three places and y gets very big or very small when x gets very big or very small or vice versa.

38

Sketch the curve with equation y = (x – 2)(x – 1)(x + 1)

General shape looking at how just the ‘x’s multiply each other, +x X +x X +x = positive x3.

---------------------------------------------------------------------------------------

Cuts through x axis when y = 0

0 = (x – 2)(x – 1)(x + 1)

x = +2 x = +1 x = -1

---------------------------------------------------------------------------------------

Cuts through y axis when x = 0

y = (0 – 2)(0 - 1)(0 + 1) = -2 X -1 X 1 = +2

---------------------------------------------------------------------------------------

As x gets very big and positive, y gets very big and positive

As x » ∞+ , y » ∞+

As x gets very big and negative, y gets very big and negative

As x » ∞- , y » ∞-

39

Sketch the curve with equation y = (x – 3)(2 - x)(x + 4)

General shape looking at how just the ‘x’s multiply each other, +x X -x X +x = negative x3.

---------------------------------------------------------------------------------------

Cuts through x axis when y = 0

0 = (x – 3)(2 - x)(x + 4)

x = +3 x = +2 x = -4

---------------------------------------------------------------------------------------

Cuts through y axis when x = 0

y = (0 – 3)(2 - 0)(0 + 4) = -3 X +2 X +4 = -24

---------------------------------------------------------------------------------------

As x gets very big and positive, y gets very big and negative

As x » ∞+ , y » ∞-

As x gets very big and negative, y gets very big and positive

As x » ∞- , y » ∞+

40

Sketch the curve with equation y = (x +2)(x – 1)2

General shape looking at how just the ‘x’s multiply each other, +x X +x X +x = positive x3.

-------------------------------------------------------------------------------------------------

Cuts through x axis when y = 0

0 = (x + 2)(x – 1)(x – 1)

x = -2 x = +1 x = +1

Notice we get the same answer twice.

-------------------------------------------------------------------------------------------------

Cuts through y axis when x = 0

y = (0 + 2)(0 - 1)(0 - 1) = +2 X -1 X -1 = +2

------------------------------------------------------------------------------------------------

As x gets very big and positive, y gets very big and positive

As x » ∞+ , y » ∞+

As x gets very big and negative, y gets very big and negative

As x » ∞- , y » ∞-

41

Sketch the curve with equation y = x3 + x2 – 2x

Start off by factorising

y = x(x2 + x – 2)

y = x(x + 2)(x – 1)

Once we have got it into brackets we can continue as normal...

Notice that any curve of this type will always go through the (0,0) point because of the x at the front.

42

We can draw transformations of straightforward y = x3 graphs.

43

And y = - x3

44

We can sketch reciprocal (y = number over x) graphs.

The number on top can be positive like y = 12x

or negative like y = - 12

x

45

We can solve quadratic equations by seeing where the graph cuts through the x axis. In the same way we can solve other equations by plotting two curves and seeing where they cross over.

The places the curves intersect is the same as the solutions. We can tell how many solutions an equation has by looking at how many times the curves cross.

46

a. On the same diagram sketch the curves y = x(x – 3) and y = x2 (1 – x).

b. Find the coordinates of the points of intersection.

The points of intersection are the same as the solutions of the equation

x(x - 3) = x2(1 – x)

If we multiply out and rearrange on one side x3 – 3x = 0 x(x2 – 3) = 0 x(x + √3)(x - √3) = 0 so x = 0, x = -√3 or x = +√3 are the three solutions.

47

We can transform a curve f(x) by a translation. Notice that which way the curve moves depends on if the number is inside or outside of the bracket.

f(x-3) = (x-3)2f(x+4) = (x+4)2 f(x) = (x)2

f(x) + 7 = (x)2 + 7

f(x) - 4 = (x)2 - 4

48

f(x) = 1/x

f(x) + 2 = 1/x + 2

f(x) - 3 = 1/x - 3

f(x-4) = 1/(x-4) f(x+5) = 1/(x+5)

The lines that the curves tend towards (but never actually reach) are called asymptotes.

49

Multiplying outside the brackets will stretch or squash the curve towards the x axis.

-2

-1

0

1

2

0 90 180 270 360

f(x) = sin x

-2

-1

0

1

2

0 90 180 270 360

½ f(x) = ½ sin x

-2

-1

0

1

2

0 90 180 270 360

2 f(x) = 2 sin x

50

Multiplying inside the brackets will stretch or squash the curve towards or away from the y axis.

-1

0

1

-360 -270 -180 -90 0 90 180 270 360

f(x) = sin x

-1

0

1

-360 -270 -180 -90 0 90 180 270 360

f(2x) = sin 2x

-1

0

1

-360 -270 -180 -90 0 90 180 270 360

f(½x) = sin ½x

51

- f(x) is a reflection in the x axis.

f(-x) is a reflection in the y axis.

Notice that – f(x) and f(-x) are not the same.

f(x)

52

Chapter 5 – Coordinate geometry in the xy plane ibvwhciwc

53

54

Chapter 6 – Sequences and Series iopwdpo

55

56

Chapter 7 - Differentiation The gradient of a line is how steep it is. We pick any two points on the line and divide the change in the y coordinates by the change in the x coordinates.

gradient = difference in y coordinatesdifference in x coordinates

-5

-4

-3

-2

-1

0

1

2

3

4

5

6

7

8

9

-3 -2 -1 0 1 2 3 4 5 6

y = 2x - 1

The gradient of a straight line is the same at any point.

A = (5, 9) B = (2, 3)

difference in ydifference in x

= 9 – 35 - 2

=

2 The gradient is 2.

57

The gradient of a curve constantly changes as we change the point we are looking at.

-5

0

5

10

15

20

25

-5 -4 -3 -2 -1 0 1 2 3 4 5

f(x) = x2

We can find out the gradient at any point on a curve by differentiating and substituting in the x value.

We normally write dydx

or f’(x) for the gradient.

58

y = x7

dydx

= 7x6

f(x) = x11

f’(x) = 11x10

y = 3x8

dydx

= 8 X

3x7 dydx

= 24x7

f(x) = 5x4

f’(x) = 4 X 5x3

f’(x) = 20x3

y = x9 + x5

dydx

= 9x8 + 5x4

f(x) = 2x5 - 3x2

f’(x) = 10x4 - 6x1

The power drops down the front and the power decreases by one.

59

y = 7x

y = 7x1

dydx

= 1 X 7x0

dydx

= 7

f(x) = 9x

f’(x) = 9

y =

6

y = 6x0

dydx

= 0 X

6x-1

f(x) = 3x5 - 7

f’(x) = 15x4

Constants disappear!

‘x’s turn into just their number.

60

y = x-3

dydx

= -3x-4

f(x) = 1x2

f(x) = x-2

f’(x) = -2x-3

y = x1/3 dydx

= 13x-2/3

f(x) = 1√x

f(x) = x1/2

f’(x) = 12x-1/2

Multiply out brackets before you differentiate.

y = x2(x – 4)

y = x3 – 4x

dydx

= 3x2 - 4

Divide out fractions before you differentiate.

f(x)= x6 – 5x

x2

f(x) =x6

x2 - 5xx2

f(x) = x4 – 5x-1

f’(x) =4x3 + 5x-2

61

We can find the gradient at any point on the curve by substituting the x value at that point.

Find the gradient of the curve y = x3 when x = 4. --------------------------------------------------------------

y = x3

dydx

= 3x2

dydx

= 3 X (4)2

dydx

= 48

Find the gradient of the curve y = 3x2 – 4x + 5 when x = 2.

--------------------------------------------------------------

y = 3x2 – 4x + 5

dydx

= 6x - 4

dydx

= 6 X 2 - 4

dydx

= 8

62

Find the gradient of the curve y = 4x2 - √x when x =

1. --------------------------------------------------------------

y = 4x2 - √x

y = 4x-2 – x1/2

dydx

= -8x-3 - 12x-1/2

dydx

= -8 X (1)-3 - 12 X (1)-1/2

dydx

= - 812

dydx

= − 172

63

We can use the gradient to find the equation of the tangent line at a point.

-5

0

5

10

15

20

25

-5 -4 -3 -2 -1 0 1 2 3 4 5

f(x) = x2

64

Find the equation of the tangent to the curve y = x3 – 3x2 + 2x - 1 at the point (3, 5).

First we are going to differentiate and substitute x = 3 to find the gradient of the curve at this point which is the same as the gradient of the tangent to the curve at this point. Then we are going to put this gradient and x1 = 3, y1 = 5 into the equation we learned in Chapter 5 and rearrange to find the equation of the line.

--------------------------------------------------------------

y = x3 – 3x2 + 2x - 1

dydx

= 3x2 – 6x + 2

dydx

= 3 X (3)2 – 6 X (3) + 2

dydx

= 11

--------------------------------------------------------------

y – y1 = m(x – x1) y – 5 = 11(x – 3) y = 11x - 28

65

We can also use the gradient to find the equation of the normal line at a point.

Remember the gradient of the normal is always minus one over the gradient of the tangent. So if the gradient of the tangent is 3 then the gradient of the

normal is − 13 .

(gradient of tangent) X (gradient of normal) = -1

gradient of normal = -1

gradient of tangent

-5

0

5

10

15

20

25

-5 -4 -3 -2 -1 0 1 2 3 4 5

f(x) = x2

66

Find the equation of the normal to the curve y = 8 - 3√x at the point x = 4.

First we are going to differentiate and substitute x = 4 to find the gradient of the curve at this point. Then we are going to calculate the gradient of the normal by working out minus one over this. Next we need to work out the value of y when x = 4 which we will find out by substituting into the original equation of the curve. And finally we are going to put the gradient, x1 and y1 into the equation we learned in Chapter 5 and rearrange to find the equation of the line.

---------------------------------------------------------------------------------------

y = 8 - 3√x y = 8 – 3x½

dydx

= − 32x-½

dydx

= − 32(4)-½

dydx

= − 34

The gradient of the tangent is − 34 so this means that the gradient of

the normal is 43 .

---------------------------------------------------------------------------------------

When x = 4 y = 8 - 3√4 y = 2 --------------------------------------------------------------------------------------- y – y1 = m (x – x1)

y – 2 = 43 (x – 4)

y = 43x -

103

67

Up till now we have been looking at dydx

as the

gradient, which it is, but another way to look at it is ‘the rate of change of y with respect to x’. Or if we increase x by ‘a bit’, how much does y increase by? We don’t just have to look at ‘y’s and ‘x’s we can use any variables.

If we draw a graph of a train setting off on a journey

we plot velocity against time. We can find dvdt

, ‘the

rate of change of velocity with respect to time’, or the acceleration.

We could look at dpdt

, ‘the rate of change of prices

with respect to time’ or inflation. We can observe a balloon that is being blown up and

calculate dVdr

, ‘the rate of change of the Volume with

respect to the radius’.

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If the velocity of a car at any time is

v = 3t2 + 7t

find dVdt

.

v = 3t2 + 7t

dVdt

= 6t + 7

If the Volume of a balloon at any time can be found using the formula

V = 43πr3

find dVdr

.

V = 43πr3

dVdr

= 4πr2

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We can differentiate twice to find the second derivative written as

d2ydx2 or f’’(x).

y = 5x3 – 4x2 + 7x - 9

dydx

= 15x2 – 8x + 7

d2ydx2 = 30x - 8

70

0

5

10

0 1 2 3

f(x) = x2

Now let’s see numerically why differentiation works. We are going to find the gradient at x = 1 for the curve y = x2.

We are going to calculate the gradient between our start point (1, 1) and our end point. Our first end point is going to be (2, 4), our next end point will be (1.5, 2.25) and the one after that (1.25, 1.5625). We will keep halving the x distance between the start point and the end point and see what happens to the gradient. The gradient of the first line is 3. The gradient of the second line is 2.5, the gradient of the third line is 2.25. If we keep halving the distance we get 2.125, 2.0625, 2.03125 as the gradient for the next three lines. Every time we halve the x distance between the start point and the end point the gradient gets closer and closer to 2. If we keep going, many. many times the x distance will eventually be practically nothing and the gradient will be practically 2.

1

2

3

4

1 2

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Now we are going to differentiate using algebra. We are going to look at two points on the graph y = x2. The start point is at (x, x2) and the end point is a tiny bit further in the x direction, we call this tiny bit extra δx (delta x). This is not a multiple, δ X x, it is all one thing, δx, the increase in x. This means the end point is at (x+δx, (x+δx)2 ). Now remember that the gradient between two lines

= difference in y coordinatesdifference in x coordinates

= end y – start yend x – start x

= (x+δx)2 – x2

x – (x + δx)

= x2 + 2xδx + δx2– x2

x – (x + δx)

= 2xδx + δx2

δx

= 2x + δx

As δx, the tiny bit extra in the x direction, tends to zero, the gradient tends to 2x.

Start point (x, x2)

End point (x+δx, (x+δx)2)

δx

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Chapter 8 - Integration Integration is the opposite of differentiating.

We say if dydx

= something then what was y equal to

before we differentiated?

We increase the power by one and divide by the new power.

If dydx

= xn then

y = xn+1

n+1

If we are just given dydx

then we never know if there

was a constant (number by itself) there before we differentiated. For example y = x2 + 7 y = x2 + 3 y = x2 - 4

will all differentiate to

dydx

= 2x

so to show we know there could have been a number there before we differentiated we add a C for constant at the end.

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f’(x) = x4

f(x) = x5

5 + C

= 15x5 + C

dydx

= x6 + x3

y = x7

7 + x4

4

= 17x7 +1

4x4 +

C

f’(x) = x2 - x5

f(x) = x3

3 - x6

6

= 13x3 - 1

6x6 +

C

dydx

= x7

dydx

= x8

8 + C

= 18x8 + C

dydx

= x-3

y = x-2

-2 + C

= − 12x-2+ C

f’(x) = x1/2

f(x) = x3/2

32

+C

= 23x3/2 + C

Remember that when we divide by a fraction we can flip it and multiply.

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When we differentiate, ‘x’s turn into numbers.

y = 3x dydx

= 3

So when we integrate, numbers turn into ‘x’s.

dy = 7 y

dydx

= 3x2 + 9

y = x3 + 9x + C

A curve has the gradient at any point of

dydx

= 3x2 + 2x

and passes through the point (2, 10). Find the equation of the curve.

------------------------------------------------------

dydx

= 3x2 + 2x

y = x3 + x2 + C

When x = 2, y = 10 so substitute to find C 10 = 3(2)3 + 2(2) + C C = - 18 The curve is y = x3 + x2 - 18

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The proper way to write integration is with the integral sign (called the summa).

∫ 3x2 dx = x3 + C The integral sign says integrate this thing.

dx does not mean d X x, it means ‘do with respect to x’. We will normally be integrating with respect to x but we will occasionally integrate with respect to other things as well.

∫ 4x3 dx = x4 + C

∫ 5x2 dx = 5x3

3 + C

∫ 7x5 + 2x4 dx = 7x6

6 + 2x5

5 + C

∫ 5t4 dt = t5 + C

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Always get rid of brackets and fractions before you integrate.

∫ (x + 1)(x + 2) dx = ∫ x2 + 3x + 2 dx

= x3

3 + 3x2

2 + 2

∫ x6 – 5x

x2 dx

= ∫ x6

x2 - 5xx3 dx

= ∫ x4 - 5x-3 dx

= x4

4 - 5x-2

- 2

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